Let $R$ and $S$ be rings and ${}_{S}A_{R}$, ${}_{R}B$, $C_{R}$, ${}_{R}D_{S}$ (bi)modules as indicated.
(i) $A\otimes_R B$ is a left $S$-module such thats $s(a\otimes b)= sa\otimes b$ for all $s\in S$, $a\in A$, $b\in B$.
Sketch of proof: (i) For each $s\in S$ the map $A\times B\to A\otimes_R B$ given by $(a,b)\mapsto sa\otimes b$ is $R$-middle linear, and therefore induces a unique group homomorphisma. $\alpha_s : A\otimes_R B\to A\otimes_R B$ such that $\alpha_s(a\otimes b)= sa\otimes b$. For each element $u=\sum_{i=1}^na_i\otimes b_i\in A\otimes_R B$ define $su$ to be the element $\alpha_s(u)=\sum_{i=1}^n \alpha_s (a_i\otimes b_i) = \sum_{i=1}^n s a_i \otimes b_i$. Since $\alpha_s$ is a homomorphism, this action of $S$ is well defined (that is, independent of how $u$ is written as a sum of generators). It is now easy to verify that $A\otimes_R B$ is a left $S$-module.
(1) Is statement of theorem precise? I mean scalar multiplication is only defined for generators of $A\otimes_R B$. By axiom of left $S$-module, we have $$s(\sum_{i=1}^na_i\otimes b_i)= \sum_{i=1}^n s(a_i\otimes b_i)= \sum_{i=1}^n s a_i\otimes b_i.$$ But well-defined of scalar multiplication on generator don’t extend over to arbitrary element of $A\otimes_R B$.
(2) Author’s proof of theorem is really cleaver; scalar multiplication is well-defined follows from $\alpha_s$ is well-defined. I am trying to prove scalar multiplication is well-defined on generators of $A\otimes_R B=F/K$ using direct approach. If $a\otimes b=c\otimes d$, then $sa\otimes b = sc\otimes d$. Suppose $a\otimes b=c\otimes d$$\iff$$(a,b)-(c,d)\in K$. We need to show $(sa,b)-(sc,d)\in K$. How to progress from here?